Question

If ${{\varepsilon }_{0}}$ and ${{\mu }_{0}}$ are, respectively, the electric permittivity and magnetic permeability of free space, $\varepsilon$ and $\mu$ are the corresponding quantities in a medium, the index of refraction of the medium in terms of the above parameters is _ _ _ _ _.
$\text{A}\text{. }\sqrt{\dfrac{\mu \varepsilon }{{{\mu }_{0}}{{\varepsilon }_{0}}}}$
$\text{B}\text{. }\sqrt{\dfrac{\varepsilon }{{{\mu }_{0}}{{\varepsilon }_{0}}}}$
$\text{C}\text{. }\sqrt{\dfrac{2}{{{\mu }_{0}}{{\varepsilon }_{0}}}}$
$\text{D}\text{. }\sqrt{\dfrac{1}{{{\mu }_{0}}{{\varepsilon }_{0}}}}$

Answer 1

The relation between the electric permittivity and the magnetic permeability of a medium is given with help of the speed (v) of light travelling in that medium as $v=\dfrac{1}{\sqrt{\mu \varepsilon }}$. In vacuum, $c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$. Refractive index is defined as $\dfrac{c}{v}$. Use this relations and find the refractive index of a medium in term of $\varepsilon$, $\mu$, ${{\varepsilon }_{0}}$ and ${{\mu }_{0}}$.

Formula used:
$v=\dfrac{1}{\sqrt{\mu \varepsilon }}$
$c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$
Refractive index = $\dfrac{c}{v}$

Complete answer:
$\varepsilon$ is called the electric permittivity of a medium. It is a proportional constant used in the formula for the electric force between two point charged particles separated by some distance, i.e. $F=\dfrac{{{q}_{1}}{{q}_{2}}}{4\pi \varepsilon {{r}^{2}}}$.
When the charges are present in the medium of vacuum, the electric permittivity is called as permittivity of free space and is denoted as ${{\varepsilon }_{0}}$.
$\mu$ is called the magnetic permeability of a medium. It is a proportional constant used in the formula for the magnetic field produced by a cuurent carrying wire of length l at a point located at a distance r, i.e. $\overrightarrow{B}=\dfrac{\mu i\left( \overrightarrow{l}\times \widehat{r} \right)}{4\pi {{r}^{2}}}$.
When the medium surrounding the wire is vacuum, the magnetic permeability is called as permeability of free space and is denoted as ${{\mu }_{0}}$.
The relation between the electric permittivity and the magnetic permeability of a medium is given with help of the speed (v) of light travelling in that medium as $v=\dfrac{1}{\sqrt{\mu \varepsilon }}$ … (i).
When we talk about vacuum, the speed of light in vacuum is equal to c. In vacuum, $\varepsilon ={{\varepsilon }_{0}}$ and $\mu ={{\mu }_{0}}$.
Therefore,
$c=\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$ …. (ii).
The refractive index or index of refraction of a medium is defined as the ratio of the speed (c) of light in vacuum to the speed (v) of light in that medium.
Hence, the refractive index is given as $\dfrac{c}{v}$.
Substitute the values of c and v from equations (i) and (ii).
Hence, the value of refractive index can be written as $\dfrac{c}{v}=\dfrac{\dfrac{1}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}}{\dfrac{1}{\sqrt{\mu \varepsilon }}}=\dfrac{\sqrt{\mu \varepsilon }}{\sqrt{{{\mu }_{0}}{{\varepsilon }_{0}}}}$.
Hence, the correct option is A.

Note:
Actually, we could have solved the question without knowing the relation between v, $\mu$ and $\varepsilon$. The only point that we must know is that the refractive index is a dimensionless quantity.
This means that out of the given options the option that has no dimension is the correct option. Option A is the square root of the ratio of two same quantities. Therefore, it has no dimensions. And option A is the only option that has no dimension.
Hence, the correct option is A.